I need to see how the phase fronts of two gaussian beams interact when they intersect with each other. The two beams are tilted towards each other so that they intersect each other at their respective minimum beam widths.

How do I perform the change of coordinate frames on one of the gaussian beams? This is what I am trying:

I pick the coordinate system of one of the beams to be (x,z) with z representing the optical axis. The origins of the two beams are a distance 'd' apart in the x dimension. I am trying to transform the coordinate system of the second beam into the first via these equations:

z$_2$ = Cos[$\alpha$] z + Sin[$\alpha$]x

x$_2$ = -Sin[$\alpha$] z + Cos[$\alpha$]x - d

(as given by the transformation matrix for rotation by $\alpha$ and accounting for the x-separation between the two beams).

I am using the usual formulation of the 2nd Gaussian beam:

\begin{equation} \frac{w_0}{w(z_2)}\text{Exp}\left[\frac{-{x_2}^2}{w(z_2)^2}\right]Exp\left[- i \text{ArcTan}\left[\frac{z_2}{z_R}\right]\right]\text{Exp}[i k \frac{{x_2}^2}{2 R(z_2)}] \end{equation}

I plugged in the values of $x_2$ and $z_2$ as given by the transformation equations given above expecting to see a tilted Gaussian beam. But the result doesn't appear to be so. I am assuming the $w_0$ (minimum spot size of the beam) and $z_R$ (the Rayleigh range of the beam) are invariant under rotations since they are lengths.

Is this method not justified? If so, what would be the proper method of handling this?


  • 2
    $\begingroup$ I don't understand the votes to close this question. It seems appropriate to me. $\endgroup$
    – garyp
    Aug 1, 2014 at 11:55
  • 1
    $\begingroup$ I haven't looked carefully, but it looks ok to me. I would suggest placing the origin at the center of the beams' shared beam waist center, and tilt both beams, one up one down. That geometry is more symmetric, which makes the math more symmetric, and would help interpreting the result. I don't know what to expect, but I wouldn't expect the result to be a Gaussian beam. $\endgroup$
    – garyp
    Aug 1, 2014 at 11:59
  • $\begingroup$ Thanks! That actually helped a lot. The rotated Gaussian beam does seem to retain most of the properties of the initial beam and looks pretty much like a rotated Gaussian. $\endgroup$ Aug 4, 2014 at 14:03


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